/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 The times of first sprinkler act... [FREE SOLUTION] | 91Ó°ÊÓ

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The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Technology, 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{sec}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.

Short Answer

Expert verified
The data contradicts the design specification; true average activation time is greater than 25 seconds.

Step by step solution

01

Define the Hypotheses

The null hypothesis is that the true average activation time, \( \mu \), is at most 25 seconds, which is \( H_0: \mu \leq 25 \). The alternative hypothesis is that the true average activation time is greater than 25 seconds, which is \( H_a: \mu > 25 \).
02

Collect Sample Data and Calculate Sample Mean

The sample data is \( \{27, 41, 22, 27, 23, 35, 30, 33, 24, 27, 28, 22, 24\} \). Calculate the sample mean, \( \bar{x} \). Sum the values and divide by the number of observations: \[ \bar{x} = \frac{27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24}{13} = 28.6154 \].
03

Calculate Sample Standard Deviation

First, calculate each deviation from the mean, square it, sum these squared deviations, and divide by \( n-1 \), where \( n \) is the number of data points. The formula for the sample standard deviation \( s \) is given by:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \].For our data:\( s \approx 5.894 \).
04

Perform the Hypothesis Test

With \( n = 13 \), calculate the test statistic using the formula: \[ t = \frac{\bar{x} - 25}{s/\sqrt{n}} \].Plugging in the values gives:\[ t = \frac{28.6154 - 25}{5.894/\sqrt{13}} \approx 2.498 \].
05

Obtain P-value

For the calculated \( t \)-value, determine the \( P \)-value using a t-distribution with \( n-1 = 12 \) degrees of freedom. From a t-table or calculator, find \( P(t > 2.498) \). The \( P \)-value \( \approx 0.014 \).
06

Make a Decision Based on P-value

Compare the \( P \)-value to the significance level \( \alpha = 0.05 \). Since \( P = 0.014 < 0.05 \), reject the null hypothesis.
07

Conclusion

The data provides sufficient evidence to conclude that the true average activation time is greater than 25 seconds, thus contradicting the design specification.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution
The t-distribution is a type of probability distribution that is used in hypothesis testing, especially when working with small sample sizes. It's similar to the normal distribution but has thicker tails. This feature allows the t-distribution to account for more variability when there's uncertainty in the sample estimate of the population standard deviation.
In our sprinkler activation time problem, we used the t-distribution because the sample size (13) is relatively small. This helps provide a more accurate estimate of the population mean, particularly when we don't know the population's standard deviation.
Key points to remember about the t-distribution include:
  • It becomes closer to the normal distribution as the sample size increases. This is because larger samples provide better estimates of the true population parameters.
  • The shape of the t-distribution is determined by the degrees of freedom, which, in this exercise, are calculated as the sample size minus one (12 degrees of freedom).
Overall, the t-distribution is crucial for making accurate inferences about populations from small sample datasets.
significance level
The significance level, denoted by \(\alpha\), is a threshold set before conducting a hypothesis test. It determines the probability of rejecting the null hypothesis when it is actually true—a scenario known as a Type I error.
In our analysis of the sprinkler activation times, we set \(\alpha = 0.05\). This means we're willing to accept a 5% risk of concluding that the true average activation time is greater than 25 seconds, even if it isn't.
Some critical points regarding significance level are:
  • A lower significance level indicates a more stringent criterion for detecting an effect or difference, reducing the likelihood of a Type I error.
  • It's essential to choose a significance level before analysis to prevent bias in hypothesis testing results.
  • Common significance levels are 0.05, 0.01, and 0.10, with 0.05 being frequently used in many fields.
Choosing the right significance level helps balance the risk of errors in statistical inference and supports the validity of the test results.
null hypothesis
The null hypothesis (\(H_0\)) is a foundational concept in hypothesis testing. It is a statement suggesting that there is no effect or no difference, and it serves as the default assumption until evidence suggests otherwise.
In our problem concerning sprinkler activation times, the null hypothesis is \(H_0: \mu \leq 25\). This implies that the true average activation time does not exceed 25 seconds.
Important considerations for the null hypothesis include:
  • It's usually formulated to reflect a statement of no change or status quo.
  • The goal of your test is to gather enough evidence to either reject or fail to reject the null hypothesis; it cannot be accepted with the same level of certainty.
  • Rejection of the null hypothesis suggests that there is significant evidence for an effect or difference, which, in our case, points towards an average activation time above 25 seconds.
Understanding the role of the null hypothesis is crucial as it guides the direction and interpretation of statistical tests.
P-value approach
The P-value approach is a method used in hypothesis testing to determine the strength of the evidence against the null hypothesis. It provides a probability measure of observing the test results, or more extreme, assuming the null hypothesis is true.
In our sprinkler activation example, we calculated a P-value of approximately 0.014. This is much lower than the significance level of 0.05.
  • When the P-value is less than the significance level, we reject the null hypothesis, indicating that there is significant evidence against \(H_0\).
  • The P-value approach allows for a nuanced understanding because it quantifies the evidence instead of a simple yes-or-no decision.
  • It's crucial to interpret the P-value in the context of the study—a smaller P-value usually signals stronger evidence against the null hypothesis.
Using the P-value approach, we concluded that the activation times are likely longer than 25 seconds, challenging the sprinkler system's design specifications.

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Most popular questions from this chapter

Let \(\mu\) denote the true average radioactivity level (picocuries per liter). The value \(5 \mathrm{pCi} / \mathrm{L}\) is considered the dividing line between safe and unsafe water. Would you recommend testing \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu>5\) or \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}\) : \(\mu<5\) ? Explain your reasoning.

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A university library ordinarily has a complete shelf inventory done once every year. Because of new shelving rules instituted the previous year, the head librarian believes it may be possible to save money by postponing the inventory. The librarian decides to select at random 1000 books from the library's collection and have them searched in a preliminary manner. If evidence indicates strongly that the true proportion of misshelved or unlocatable books is less than \(.02\), then the inventory will be postponed. a. Among the 1000 books searched, 15 were misshelved or unlocatable. Test the relevant hypotheses and advise the librarian what to do (use \(\alpha=.05\) ). b. If the true proportion of misshelved and lost books is actually .01, what is the probability that the inventory will be (unnecessarily) taken? c. If the true proportion is \(.05\), what is the probability that the inventory will be postponed?

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